Continuous random variables with joint density function

Question

Given that X, Y are continuous random variables with joint density function

$$f_{x,y}=x-y+1$$ And: $$0 \leq x \leq 1$$ $$0 \leq y \leq 1$$

Need to calculate this: $$P(y\geq \frac{1}{4}|x=\frac{1}{4})$$

How can I prove that? I tried this: $$\frac{P(y\geq \frac{1}{4},x=\frac{1}{4})}{P(x=\frac{1}{4})}$$

And after that I draw the domain (desmos.com):

finally I tried to make Integral and I got zero...

Answer 1

We will need the conditional density of $Y$ given $X$. Notice that $$f_X(x) = \int_0^1 f_{X,Y}(x,y)\,dy = x+\frac{1}{2}.$$

Thus, we have that $$f_{Y|X}(y|x) = \frac{f_{X,Y}(x,y)}{f_X(x)} = \frac{x-y+1}{x+1/2}.$$

Finally, we have $$P(Y\geq 1/4|X = 1/4) = \int_{1/4}^1 f_{Y|X}(y|1/4)\,dy = \int_{1/4}^1\frac{1/4-y+1}{1/4+1/2}\,dy = \frac{5}{8}$$

Answer 2

Let's use definitions. By definition, $$ \mathbb{P}\left.\left(Y\geq\frac{1}{4}\right|X=\frac{1}{4}\right)=\int_{1/4}^1 f_{Y|X}(y|1/4)dy, $$ where $f_{Y|X}$ is the conditional PDF, which is (by definition) $$ f_{Y|X}(y|x)=\frac{f_{X,Y}(x,y)}{f_{X}(x)}, $$ where $f_Y$ is the marginal PDF of $Y$, which is (by definition) $$ f_X(x)=\int_0^1 f_{X,Y}(x,y)dy. $$ Please complete the remaining integral calculations yourself.